MECHANICS OF SOLIDS ME F211 BITS Pilani K K Birla Goa Campus VIKAS CHAUDHARI Mechanics of Solids

Chapter-5 -Strain-Temperature Relations

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Contents:  Introduction  Tensile test  Stress- Strain diagram  Idealizations of Stress-Strain Curves  Elastic Stress- Strain Relations  Thermal Strain  Complete Equations of Elasticity  Strain Energy in an Elastic Body  Criteria for Initial Yielding

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Introduction Elastic : It is the deformation which exist when load applied and it disappears as soon as the load is released. Plastic Deformation: when load is applied beyond elastic range, and the deformation does not disappears after the load is released. A Ductile materials is one which the plastic deformation is much larger than elastic deformation

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Tensile Test

Tensile Test

Displacements in Tensile Test

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Stress- Strain diagram Stress-strain diagram for a structural steel in tension

Not to scale Draw to scale

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Stress- Strain diagram Typical stress- strain diagram for an aluminum alloy

Offset yield point (proof stress)

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Stress- Strain diagram Typical stress-strain diagram for brittle material

Stress-strain diagram for a cast iron in tension and compression

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Stress- Strain diagram Loading, Unoading and Reloading

(a) Elastic behaviour Reloading of a material and (b) Partially elastic behaviour raising of the yield stress

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Idealizations of Stress-Strain Curves

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Example: Two Coaxial tubes, the inner one of 1020 CR steel and cross sectional area

As, and the outer one of 2024-T4 aluminum alloy and of area Aa are compressed between heavy , flat end plates as shown in figure. We wish to determine the load-deflection curve of the assembly as it is compressed into the plastic region by an axial force P. The idealized stress strain curves are shown for both the tube materials in the second figure

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Solution In the range 0 0.0032,

sEEEE s  s  s  and  a  a  a  a  590 where E 184 GPa s 0.0032 380 E 76 GPa a 0.005

In the range 0.0032 0.005,

sY s 590 MPa and  a  E a  a  E a 

In the range 0.005  ,

sY s 590 MPa and a  Y a  380 MPa

Equilibrium

FAAPy  s s  s a   0

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Elastic Stress- Strain Relations  Assuming that the material is isotropic and homogeneous.  Consider an element on which there is only one component of normal stress acting, as shown in figure.  For linear elastic solid, stress is directly proportional to strain.  Uniaxial tension (compression) test shows that lateral compressive (extensional) strain    x is a fixed fraction of longitudinal extensional x E (compressive) strain.          x  This fixed fraction is known as Poisson’s ratio y z x E and given by symbol .

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Elastic Stress- Strain Relations  For isotropic material normal stress does not produce shear strain.

Similarly only normal stress  y acts Analogous results are

yyobtained for the strain y and  x   z    y    EEdue to z.

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Elastic Stress- Strain Relations  For isotropic material does not produce normal strain.  Each shear-stress component produces only its corresponding shear- strain component.  For isotropic materials, shear stress is directly proportional to the shear strain. The proportionality constant is G ().

 xy  yz  zx xyGGG yz zx

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Elastic Stress- Strain Relations

1        xE  x y z  1 y  y    x   z  E  1        zE  z x y   xy  yz  zx xyGGG yz zx Relation between E, G and  11      and   IEE II  21      xyGE I II E  2G 1 

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Thermal Strain  The effect of temperature on elastic constants for many materials is small for temperature change of a hundred degree centigrade and we will not consider for our study  The strain due to temperature change in the absence of stress is called thermal strain and it is denoted by superscript t on strain symbol (t)  For isotropic material, thermal strain is pure expansion or contraction with no shear components.

 Thermal strain due to a change in temperature from To to T is

t t t x  y   z   TT  o 

t t  t  0 xy yz xz  The quantity  is coefficient of linear expansion.

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Thermal Strain  Denoting the elastic strain due to stress by e and the thermal part by t, the total strain is

et   

 If material is rigidly restrained so that no strain is possible, the elastic part of the strain will be the negative of the thermal strain.

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Complete Equations of Elasticity  Equilibrium Equations  Geometric Compatibility  Stress-strain Temperature Relations

1  x  x    y   z   TT  o  E   1 y  y    x   z   TT  o  E  1  z  z    x   y   TT  o  E    xy  yz  zx xyGGG yz zx

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Example A long, thin plate of width b, thickness t, and length L is placed between two rigid walls a distance b apart and is acted on by an axial force P, as shown in figure. We wish to find the deflection of the plate parallel to the force P.

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Solution Assumptions 1. The axial force P results in an axial normal stress uniformly distributed over the plate area, including the end areas. 2. There is no normal stress in the thin direction (plane stress in x-y plane).

3. There is no deformation in the y direction, that is, y = 0 (plane strain in x-z plane). 4. There is no friction force at the walls. 5. The normal stress of contact between the plate and wall is uniform over the length and width of the plate.

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Solution

Equilibrium P         0 xbt y o z

xy yz  xz  0

Geometric Compatibility  0 and   yxL Stress Strain Relations 11              xEEE x y y y x z y x 

xy yz  xz  0

After solving

P2 PL P  y  x      1    z   1     bt  Ebt Ebt1 L

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Problem The stress in a flat steel plate in a condition of plane stress are,

x = 130MPa; y = -70MPa and xy = 80MPa. Find the magnitude and orientation of the principal strains in the plane of the plate. Find also the magnitudes of the third principal strain

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Problem In a flat steel plate which is loaded in the xy plane, it is known that -4 x = 145MPa; xy = 42MPa and z = -3.610 .

What is the value of the stress y?

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Problem A steel pipe is held by two fixed supports as shown in figure. When mounted, the temperature of the pipe was 20oC. In use, however, cold fluid moves through the pipe, causing it to cool considerably. If we assume that the pipe has uniform temperature of -15oC and if we take the coefficient of linear expansion to be 1210-6/oC for this temperature range, determine the state of stress and strain in the central portion of the pipe as a result of this cooling. Neglect the local end effects near the supports and neglect body forces and fluid pressure and drag forces.

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Problem It is desired to produce a tight fit of a steel shaft in a steel pulley. The internal diameter of the hole in the pulley is 24.950mm, while the outside diameter of the shaft is 25.000mm. The pulley will be assembled on the shaft by either heating the pulley or cooling the shaft and then putting the shaft in the pulley hole and allowing the assembly to reach uniform temperature. Is it more effective to heat the pulley or to cool the shaft? What temperature change would be required in each case to produce a clearance of 0.0025 mm for easy assembly?

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Problem A circular bar (E = 200GPa,  = 0.32, and  = 11.710-6 /oC) has a diameter of 100mm. The bar is built into a rigid wall on the left, and a gap of 0.5 mm exists between the right wall and the bar prior to an increase in temperature as shown in figure. Temperature of the bar is increased uniformly by 80oC. Determine the average axial stress and the change in the diameter of the bar.

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Strain Energy in an Elastic Body The energy stored in a body due to deformation is called strain energy. For small element shown in figure (a),

11 dUx dydz  x dx  x  x dV 22 1 1P   1 U  dV  dV  P  xx    2VV 2AL   2 Similarly considered small element shown in figure (c),

11 dU dxdz  dy   dV 22 xy xy xy xy

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Criteria for Initial Yielding Mises yield criterion According to Mises criterion, yielding occurs in three-dimensional state of stress when the root mean square of the differences between the principal stresses reaches the same value that it has when yielding occurs in a tensile test.

12 2 2  1  2 2 2  2 1  2   2  3   3  1  YYY 0  0 0  0  3  3   3

1 2 2 2 1  2   2   3   3   1   Y 2  It is also known as distortion- energy criterion or octahedral shear stress criterion for yielding

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Criteria for Initial Yielding Tresca yield criterion According to Tresca criterion, the yielding occurs when the absolute maximum shear stress at a point reaches the value of the maximum shear stress to cause yielding in a tensile test

 Y  max min max 22

It is also known as maximum shear stress criterion.

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Criteria for Initial Yielding Geometrical representation in principal stress space of the Mises and Tresca yield criteria.

Plane stress case

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Problem A batch of 2024-T4 aluminum alloy yields in uniaxial tension at

stress o = 330MPa. If this material is subjected to the following state of stress, will it yield according to a. The Mises criterion b. Tresca criterion

x = 138MPa xy = 138MPa

y = -69MPa yz = 0

z = 0 xz = 0

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References 1. Introduction to Mechanics of Solids by S. H. Crandall et al (In SI units), McGraw-Hill

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